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A032169
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Number of aperiodic necklaces of n beads of 2 colors, 11 of them black.
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4
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1, 6, 26, 91, 273, 728, 1768, 3978, 8398, 16796, 32065, 58786, 104006, 178296, 297160, 482885, 766935, 1193010, 1820910, 2731365, 4032015, 5864749, 8414640, 11920740, 16689036, 23107896, 31666376, 42975796
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OFFSET
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12,2
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COMMENTS
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Assume n >= k >= 2. If a_k(n) is the number of aperiodic necklaces of n beads of 2 colors such that k of them are black and n-k of them are white, then a_k(n) = (1/k)*Sum_{d|gcd(n,k)} mu(d)*binomial(n/d - 1, k/d - 1) = (1/n)*Sum_{d|gcd(n,k)} mu(d)*binomial(n/d, k/d). This follows from Herbert Kociemba's general formula for the g.f. of (a_k(n): n>=1) that can be found in the comments for sequence A032168.
For k prime, we get a_k(n) = floor(binomial(n-1, k-1)/k). In such a case, the sequence becomes a column for triangle A011847. (This is not true when k is composite >= 4.)
(End)
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (10, -45, 120, -210, 252, -210, 120, -45, 10, -1, 1, -10, 45, -120, 210, -252, 210, -120, 45, -10, 1).
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FORMULA
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"CHK[ 11 ]" (necklace, identity, unlabeled, 11 parts) transform of 1, 1, 1, 1, ...
a(n) = (1/11)*(binomial(n-1, 10) - I(11|n)) = floor(binomial(n-1, 10)/11) for n >= 12, where I(a|b) = 1 if integer a divides integer b, and 0 otherwise. - Petros Hadjicostas, Aug 26 2018
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MATHEMATICA
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CoefficientList[Series[x^11/11 (1/(1-x)^11-1/(1- x^11)), {x, 0, 50}], x] (* Herbert Kociemba, Oct 16 2016 *)
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CROSSREFS
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KEYWORD
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nonn,changed
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AUTHOR
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STATUS
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approved
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